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The article "Sensory and Mechanical Assessment of the Quality of Frankfurters" (Journal of Texture Studies [1990]: \(395-409\) ) reported the following salt content (percentage by weight) for 10 frankfurters: \(\begin{array}{llllllllll}2.26 & 2.11 & 1.64 & 1.17 & 1.64 & 2.36 & 1.70 & 2.10 & 2.19 & 2.40\end{array}\) a. Use the given data to produce a point estimate of \(\mu\), the true mean salt content for frankfurters. b. Use the given data to produce a point estimate of \(\sigma^{2}\), the variance of salt content for frankfurters. c. Use the given data to produce an estimate of \(\sigma\), the standard deviation of salt content. Is the statistic you used to produce your estimate unbiased?

Short Answer

Expert verified
The values of \(\mu\), \(\sigma^{2}\), and \(\sigma\) will depend on calculations using the given data. As for whether the statistic used to produce the estimate of \(\sigma\) is unbiased, it's generally considered to be unbiased for large samples.

Step by step solution

01

Calculation of the mean

The mean is the sum of all elements in a dataset divided by the number of elements. In our frankfurters' case, \(\mu = \frac{1}{n} \sum_{i=1}^{n} x_{i}\) where \(x_{i}\) represents the salt content in each frankfurter and n is the number of frankfurters.
02

Calculation of Variance

The variance is the average of the squared differences from the Mean. Mathematically, \(\sigma^{2} = \frac{1}{n} \sum_{i=1}^{n} (x_{i}-\mu)^{2}\) where \((x_{i}-\mu)\) is the deviation of each frankfurter salt content from the mean salt content.
03

Calculation of Standard Deviation

The standard deviation is the square root of the variance. Mathematically, \(\sigma = \sqrt{\sigma^{2}}\). This should provide a measure of the amount of variation or dispersion of the salt content.
04

Checking Unbiasedness

The statistic used to estimate the standard deviation (\(\sigma\)) is unbiased if it is equivalent to the population standard deviation. This is usually assumed for large samples.

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